I can do it for you if it just come to the colors inside the circles (It will take me 30 minutes) but how could I replace the links (in yellow) between circles?After rethinking I believe that it will hard to guess the rules.So if no one find the rules (which is part of the puzzle) then I will post the rules.Why I said new kind of puzzle? Because all the puzzles are presented with rules, diagrams and so on.A picture could often replace words.

houlahop wrote:I can do it for you if it just come to the colors inside the circles (It will take me 30 minutes) but how could I replace the links (in yellow) between circles?After rethinking I believe that it will hard to guess the rules.So if no one find the rules (which is part of the puzzle) then I will post the rules.Why I said new kind of puzzle? Because all the puzzles are presented with rules, diagrams and so on.A picture could often replace words.

If all the links (that are not part of the "grid") are the same color, then leave them as is. If there is more than one type of link, consider something like dashed lines, or double lines(like =) which will allow you 5 different types of links (single solid line, single dashed line, double solid line, double dashed line, one solid line with one dashed line)

Here are the 2 boards with circles. I replaced the colors by letters (4 colors 4 letters the choice was arbitrary so B is not blue for example)As you can see the links then there is no problem.Good luck!

Are the rules just straight lines only between different colors and the number of intersections each line has? It looks like there are 2 lines each with 0, 1, 2, and 3 intersections with other lines. If that's a rule, then two of the lines in the big box will need to make 15 intersections?

"Are the rules just straight lines only between different colors..."That is the first rule : straight line between 2 different colors. That means that it is not allowed to link circles having the same color.

" and the number of intersections each line has? It looks like there are 2 lines each with 0, 1, 2, and 3 intersections with other lines. If that's a rule, "A player in the case of board 4x4 must use exactly 2 lines of each length (1,r(2),r(5),r(10))

" then two of the lines in the big box will need to make 15 intersections?"The previous rule apllied to the board 8x8 implies that a player must use 8 lines of each length (1,r(2),r(5),r(10)).

Sorry I did not understand the use of "intersections".Anyway it seems that you have decrypted the rules.Some questions remain :- Does the board 8x8 have at least one solution?- If we generate randomly a board 8x8 with 16 circles of each of the 4 colors are we sure to find at least one solution?- Can we find an efficient algorithm to solve quickly any configuration 8x8 board?- Finding solution seems to be very hard but there is maybe a strategy on how to attack the puzzle and solve it (it is a conjecture from my part)

houlahop wrote:You wrote :" then two of the lines in the big box will need to make 15 intersections?"The previous rule applied to the board 8x8 implies that a player must use 8 lines of each length (1,r(2),r(5),r(10)).

Ah ok. Now I understand what you were talking about.Intersections between the links (straight lines) does not matter at all.What is important is that all circles be connected by pairs : one straight line linking 2 circles.Player must use 8 straight lines of each length.8*2*4=64 circles.Now we need to solve the puzzle 8x8.

Are we talking links of r(n) for each n=1²+[0..max_orthagonal]², then? For 4x4, that was r(1+0),r(1+1),r(1+4),r(1+9), four lengths (otherwise codable as {0,1} {1,1} {2,1} and {3,1}) 'reserving' two points each, which is half of the 16, thus two of each length.

With 8x8, there's {0,1}..{7,1}, for eight distinct lengths, 16 points in that batch of singletons, and thus we need four of each length?

That would be most logical, given then lack of {2,2} (or other {>1,≠1} pairings) in the original, but having already lost ground after identifying Green as the common factor with {0,1} links (or opposite in orthogonality), Red with {2,1} and Blue with {3,1}, but Yellow apparently not common to both {1,1}s, I perhaps ought to not make assumptions.

(As I now understand it, the colours are not only not "Green=1", throughout the whole puzzlescape, however arbitrarily decided, but not even within the same puzzle... The colours are merely there to constrain "never from any colour to the same colour", and beyond that it's just a kind of beyond-non-euclidean Four Colour Problem...)

What are invariants no matter the board size :- 4 colors (blue, red, yellow, green) - 4 lengths (1,r(2),r(5),r(10). Easy to compute the number needed. Number of straight lines used is equal. In case 4x4 thi number is equal to 2. In case 8x8, this number is equal to 8. - linking all the circles 2 by 2 with different colors. You are not allowed to connect 2 circles having the same color.

In fact it is hard to deduce the rules starting from one picture. I was not aware of that because the are other options you could derive from the picture.

Thank you for your interest and your comments.Picking randomly the colors will lead each time to an open puzzle. Does it hav at least one solution or no.I tried to build a configuration with no solution at all (4x4) but I did not finish my job.

If we generate randomly a board 8x8 with 16 circles of each of the 4 colors are we sure to find at least one solution?

Clearly not. The rules for line length give us at most 24 possible targets from a single circle. That's not so bad, but circles near an edge have less - in the most extreme case, a corner only has 7. With 16 of each color, you could randomly end up withowithout any possible connection, let alone one that can be used without cutting off other circle's needs. For larger boards you need to allow longer lines instead of just using more lines of each length if you want to avoid this issue. But I suppose that could get visually messy.

In board 4x4 once we find one solution we could generate 2^8 solutions minus 1 (255 solutions) by switching the colors. We assume that the links (straight lines)remain as they are.So there is a way (messy for sure) to compute at least (in terms of percentage) the number of solutions out of 4^16.To those who try to solve it is this puzzle boring or exciting?

There is a way to solve this 8x8 in particular.If you break the board 8x8 into 4 boards 4x4 you will find easily the solution.It is not a general method to find at least one solution.I still think that there is some method or algorithm to solve any 8x8 board quickly.I will post the solution if no one found it.